Calculus - Basic problems 2

Solution 1

1. Draw the graph and the tangent at $x=2$ (see below). Based on the figure, the slope of the tangent is $$a_t=\frac{\Delta y}{\Delta x}\approx \frac{1}{3}$$
2. We approximate the tangent by a secant by moving to the right by $h=0.1$ (see figure). The slope of the secant is an approximation of the slope of the tangent (the smaller the displacement $h$, the better is the approximation). Using the slope triangle, we get for the slope of the secant $$a_s=\frac{\Delta y}{\Delta x}=\frac{f(2.1)-f(2)}{0.1}$$ With $f(2)=1.58740$ and $f(2.1)=1.63988$, we get $$a_s=\frac{1.63988-1.58740}{0.1}=0.5248$$
3. The exact slope can be calculated using the rules of differentiation. With $f(x)=x^{2/3}$, we get $$f'(x)=\frac{2}{3}x^{-1/3}$$ and thus $$a_t=f'(2)=\frac{2}{3}\cdot 2^{-1/3}=0.5291$$
4. Using three bars, the width of each bar has to be $\Delta x=\frac{2}{3}$, and thus we get for the three bar areas: $$A_1=\frac{2}{3}\cdot f(\frac{2}{3})=\frac{2}{3}\cdot \left(\frac{2}{3}\right)^{2/3}=0.508$$ $$A_2=\frac{2}{3}\cdot f(\frac{4}{3})=\frac{2}{3}\cdot \left(\frac{4}{3}\right)^{2/3}=0.807$$ $$A_3=\frac{2}{3}\cdot f(\frac{6}{3})=\frac{2}{3}\cdot 2^{2/3}=1.058$$ Thus, an approximation of the area $A$ under the curve is $$A\approx A_1+A_2+A_3 = 2.374$$
5. Using the calculator, we get for the area under the curve: $$A\approx 1.9048...$$
6. The exact value can be calculated using the fundamental theorem of calculus. The antiderivative of $f(x)=x^{2/3}$ is $$F(x)=\frac{3}{5}x^{5/3}$$ and thus $$\begin{array}{lll} A = \int_0^2 f(x)\,dx &=& F(2)-F(0)\\ &=& \frac{3}{5}\cdot 2^{5/3} - \frac{3}{5}\cdot 0^{5/3}\\ &=& 1.9048... \end{array}$$ Note that the numerical estimate of the integral using the calculator (see 5) is normally really close to the exact value.

Solution 2

The derivative of $f(x)=x^3$ is $f'(x)=3x^2$, and the antiderivative is $F(x)=\frac{1}{4}x^4$.

1. The slope of the tangent at $x=2$ is $f'(2)=3\cdot 2^2=12$
2. The angle is (SOHCAHTOA, see graph at the bottom, left): $$\tan(\alpha)=\frac{O}{A}=\underbrace{\frac{\Delta y}{\Delta x}}_{\text{slope of tangent}}=12$$ Thus, $$\alpha=\arctan(12)=\tan^{-1}(12)=85.23^\circ$$
3. The equation of the tangent is a linear function: $t(x)=ax+b$ where the slope $a=12$. Thus, it is $$t(x)=12x+b$$ To find $b$ (the $y$-intercept), we use the fact, that the tangent touches the graph of $f$ at $x=2$, that is, we have $$t(2)=f(2)$$ and thus $$12\cdot 2+b=2^3=8 \rightarrow b=-16$$ Thus, the equation of the tangent is $t(x)=12x-16$.
4. Find $x$ with $$f'(x)=9 \rightarrow 3x^2=9$$ It follows $x=\pm \sqrt{3}$. Thus, there are two tangents to the graph of $f$ with slope $9$. One is at $P_1(-1.732|-5.196)$ and $P_2(1.732|5.196)$.
5. We have $$\int_{-1}^2 x^3\, dx = F(2)-F(-1)=\frac{1}{4}\cdot 2^4 - \frac{1}{4}\cdot(-1)^4=3.75$$
6. The integral gives the signed area. To find the area, we have to determine the area below the $x$-axis and the one above the $x$-axis individually (see figure below, right): $$A_1=\int_{-1}^0 x^3\, dx = F(0)-F(-1)=\frac{1}{4}\cdot 0^4 - \frac{1}{4}\cdot(-1)^4=-0.25$$ and $$A_2=\int_{0}^2 x^3\, dx = F(2)-F(0)=\frac{1}{4}\cdot 2^4 - \frac{1}{4}\cdot 0^4=4$$ Thus, the area is $A=0.25+4=4.25$.

Solution 3

See the figure below. At $x=0$ there is no single tangent to the graph of $f$. Thus, in this case it is not clear what $f'(0)$ show be. We say that $f$ is not differentiable at $x=0$. Everywhere else we can determine $f'(x)$.

Solution 4

1. The four basic rules are
   1. Power rule: $$f(x)=x^n \rightarrow f'(x)=n x^{n-1}$$
   2. Sum rule: $$f(x)=u(x)+v(x) \rightarrow f'(x)=u'(x)+v'(x)$$ or with coefficients $a$ and $b$: $$f(x)=a \cdot u(x)+ b \cdot v(x) \rightarrow f'(x)=a \cdot u'(x)+b \cdot v'(x)$$
   3. product rule: $$f(x)=u(x)\cdot v(x) \rightarrow f'(x)=u'(x)\cdot v(x)+u(x)\cdot v'(x)$$
   4. chain rule: $$f(x)=u({\color{red} v(x)}) \rightarrow f'(x)=u'({\color{red} v(x)})\cdot v'(x)$$
2. It is
   1. $f(x)=x^2+3x-1$ $\rightarrow f'(x)=2x+3$
   2. $f(x)=3 =3x^0$ $\rightarrow f'(x)=0$
   3. $f(x)=x\sqrt{x}=x^{3/2}$ $\rightarrow f'(x)=\frac{3}{2} x^{1/2}=1.5\sqrt{x}$
   4. $f(x)=3x^5+4x^2-2\sqrt{x}$ $\rightarrow f'(x)=15x^4+8x-x^{-1/2}$
   5. $f(x)=x^2(1-x)=x^2-x^3$ $\rightarrow f'(x)=2x-3x^2$
   6. $f(x)=\sqrt{x}(x^2-2x+1)=x^{2.5}-2x^{1.5}+x^{0.5}$ $\rightarrow f'(x)=2.5x^{1.5}-3x^{0.5}+0.5x^{-0.5}$
   7. $f(x)=2(x-1)^2 =2x^2-4x+2$ $\rightarrow f'(x)=4x-4$
   8. $f(x)=2({\color{red}x-1})^5$ $\rightarrow f'(x)=10({\color{red}x-1})^4\cdot 1=10(x-1)^4$ (chain rule)
   9. $f(x)=2({\color{red}1-2x})^5$ $\rightarrow f'(x)=10({\color{red}1-2x})^4\cdot (-2)=-20(1-2x)^4$ (chain rule)
   10. $f(x)=(2x-1)(x+1)$ $\rightarrow f'(x)=2\cdot(x+1)+(2x-1)\cdot 1=4x+1$ (product rule) or $f(x)=(2x-1)(x+1)=2x^2+x-1$ $\rightarrow f'(x)=4x+1$
   11. $f(x)=2(x-1)x(x+3)=2x^3+4x^2-6x$ $\rightarrow f'(x)=6x^2+8x-6$
   12. $f(x)=\frac{2}{x}+\frac{3}{\sqrt{x}}+\frac{4}{x^2} = 2x^{-1}+3x^{-0.5}+4x^{-2}$ $\rightarrow f'(x)=-2x^{-2}-1.5x^{-1.5}-8x^{-3}=-\frac{2}{x^2}-\frac{1.5}{x^{1.5}}-\frac{8}{x^3}$
   13. $f(x)=\ln(x)\cdot \sin(x)$ $\rightarrow f'(x)=\frac{1}{x}\cdot \sin(x)+\ln(x)\cdot \cos(x)$ (product rule)
   14. $f(x)= x^2 \cdot e^x$ $\rightarrow f'(x)=2x e^x+x^2e^x=e^x x(2+x)$ (product rule)
   15. $f(x)= \cos({\color{red}x^3})$ $\rightarrow f'(x)=-\sin({\color{red}x^3})\cdot 3x^2$ (chain rule)
   16. $f(x)= \ln({\color{red}\sin(x)})$ $\rightarrow f'(x)=\frac{1}{{\color{red}\sin(x)}}\cdot \cos(x)$ (chain rule)
   17. $f(x)=\sqrt{{\color{red}1+x^2}}=({\color{red}1+x^2})^{0.5}$ $\rightarrow f'(x)=0.5({\color{red}1+x^2})^{-0.5}\cdot 2x=\frac{x}{\sqrt{1+x^2}}$ (chain rule)
   18. $f(x)=({\color{red}\sin(x)})^2$ $\rightarrow f'(x)=2{\color{red}\sin(x)}\cdot \cos(x)$ (chain rule). You can also use the product rule ... .
   19. $f(x)=(x e^x)^3=x^3 e^{{\color{red}3x}}$ $\rightarrow f'(x)=3x^2 \cdot e^{3x}+x^3\cdot e^{{\color{red}3x}}\cdot 3=3e^{3x}(x^2+x^3)$ (product rule and chain rule)
   20. $f(x)=\frac{1}{\sqrt{1+x^2}}=({\color{red}1+x^2})^{-0.5}$ $\rightarrow f'(x)=-0.5({\color{red}1+x^2})^{-1.5}\cdot 2x = -\frac{x}{(1+x^2)^{1.5}}$ (chain rule)
   21. $f(x)=(2x^3+15x+1)\cdot e^{{\color{red}-2x^2}}$ $\rightarrow f'(x)=6x^2+15)e^{-2x^2}+(2x^3+15x+1)\cdot e^{{\color{red}-2x^2}}\cdot (-4x)$ (product rule and chain rule)
   22. $f(x)=(x^4-2x^3)\cdot\sin({\color{red}2x^2})$ $\rightarrow f'(x)=(4x^3-6x^2)\sin(2x^2)+(x^4-2x^3)\cdot\cos({\color{red}2x^2})\cdot 4x$ (product rule and chain rule)

Solution 5

Let's denote by $F$ the antiderivative of $f$.

1. $f(x)=x^n \rightarrow F(x)=\frac{1}{n+1} x^{n+1}$
2. It is
   1. $f(x)=5=5x^0$ $\rightarrow F(x)=5\cdot \frac{1}{1}x^1=5x$
   2. $f(x)=5x = 5x^1$ $\rightarrow F(x)=5\cdot \frac{1}{2}x^2=$
   3. $f(x)=5x^2$ $\rightarrow F(x)=5\cdot \frac{1}{3}x^3=\frac{5}{3}x^3$
   4. $f(x)=5\sqrt{x}=5x^{0.5}$ $\rightarrow F(x)=5\cdot \frac{1}{1.5}x^{1.5}=\frac{10}{3}x^{1.5}$
   5. $f(x)=\frac{5}{x}$ $\rightarrow F(x)=5\ln(x)$
   6. $f(x)=\frac{5}{x^2}=5x^{-2}$ $\rightarrow F(x)=5\cdot \frac{1}{-1}x^{-1}=-\frac{5}{x}$
   7. $f(x)=\frac{5}{\sqrt[3]{x^4}}=5x^{-4/3}$ $\rightarrow F(x)=5\cdot \frac{1}{-1/3}x^{-1/3}=-\frac{15}{\sqrt[3]{x}}$
   8. $f(x)=\frac{5}{x}+2$ $\rightarrow F(x)=5\cdot \ln(x)+2x$
3. The fundamental theorem of calculus: $\int_a^b f(x)\, dx=F(b)-F(a)$
   1. $F(x)=\frac{1}{2}x^2$ $\rightarrow \int_1^2 x\, dx = (\frac{1}{2}\cdot 2^2)-(\frac{1}{2}\cdot 1^2)= 1.5$
   2. $F(x)= \frac{2}{3}x^3+1.5x^2+x$ $\rightarrow \int_0^1 (2x^2+3x+1)\, dx = (\frac{2}{3}\cdot 1^3+1.5\cdot 1^2+1) - (\frac{2}{3}\cdot 0^3+1.5\cdot 0^2+0)=3.1\overline{6}$
   3. $F(x)=-2\cos(x)$ $\rightarrow \int_{\pi/2}^\pi 2\sin(x)\, dx=(-2\cos(\pi))-(-2\cos(\pi/2))=2$
   4. $F(x)= 1.2x^{2.5}$ $\rightarrow \int_1^2 3x\sqrt{x}\, dx=1.2\cdot 2^{2.5}-1.2\cdot 1^{2.5}=5.588$
   5. $F(x)=\frac{1}{6}x^3-\frac{1}{8}x^4$ $\rightarrow \int_{-1}^1 \frac{1}{2} x^2(1-x)\, dx = \left(\frac{1}{6}\cdot 1^3-\frac{1}{8}\cdot 1^4\right)-\left(\frac{1}{6}\cdot (-1)^3-\frac{1}{8}\cdot (-1)^4\right)=\frac{1}{3}$
   6. $F(x)=2\ln(x)+x$ $\rightarrow \int_1^e (\frac{2}{x}+1)\, dx= (2\ln(e)+e)-(2\ln(1)+1)=e+1$

Solution 6

1. It is
   1. A point $P$ on the graph of $f$ where the tangent is horizontal is called a stationary point of $f$. It is a local maximum if it is the top of the hill, and a local minimum if it the bottom of a valley. It is a saddle point if it looks like a saddle, that is, it goes downwards if moving to the left and upwards if moving to the right (or vice versa). In the figure below, $A$ is a saddle point, $B$ is a local maximum and $C$ a local minimum.
   2. A point $P$ on the graph of $f$ where the slope of the tangent reaches its locally highest (or locally lowest) slope is called an inflection point of $f$. Saddle points are inflection points and stationary points. In general, inflection points do not have to be stationary points. In the graph below, points $A$, $D$ and $E$ are inflection points.
2. From the sketch of the first derivative, we see the following for a point $P(x|y)$ on the graph of $f$:
   1. $f'(x)=0, f''(x)>0 \rightarrow P\text{ local min}$
   2. $f'(x)=0, f''(x)<0 \rightarrow P\text{ local max}$
   3. $f'(x)=0, f''(x)=0, f'''(x)\neq 0 \rightarrow P\text{ inflection pt and saddle pt}$
   4. $f'(x)\neq 0, f''(x)=0, f'''(x)\neq 0 \rightarrow P\text{ inflection pt, no saddle pt}$ If $f'(x)=f''(x)=f'''(x)=0$ it might be a flat saddle point, local maxima or local minima. Draw the graph to decide.
3. $g(x)=x^3(1-2x)=x^3-2x^4$. $g'(x)=3x^2-8x^3$, $g''(x)=6x-24x^2$, $g'''(x)=6-48x$.
   1. stationary points: Find $x$ with $$g'(x)=0 \rightarrow 3x^2-8x^3=x^2(3-8x)=0$$ thus $x_1=0$ and $x_2=\frac{3}{8}$. The stationary points are $P_1(0|0)$ and $P_2(0.375|0.0131)$.
   2. local max/min/saddle pt? $g''(0)=0$ so $P_1$ is not a local extrema. Because $g'''(0)=6\neq$ we see that $P_1$ is a saddle point. $g''\left(\frac{3}{8}\right)=-\frac{9}{8}<0$. Thus, $P_2$ is a local max.
   3. Inflection points? Find $x$ with $$g''(x)=0 \rightarrow x(6-24x)=0$$ thus $x_1=0$ and $x_2=0.25$. The point at $x_1$ is a saddle point (see above). The point at $x_2$ is also an inflection point, because $g'''(0.25)=-6\neq 0$. Its coordinates are $P_3(0.25|0.0078)$.